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CBSE CLASS X – MATHEMATICS (STANDARD) SAMPLE QUESTION PAPER
SET – 4 | ACADEMIC SESSION 2026–2027
General Instructions:
- This question paper contains 38 questions divided into 5 Sections: A, B, C, D and E.
- All Questions are compulsory.
- Section A comprises 20 Multiple Choice Questions (MCQs) of 1 mark each (Q1 to Q18 are MCQs and Q19 & Q20 are Assertion–Reason based).
- Section B comprises 5 Very Short Answer (VSA) questions of 2 marks each (Q21 to Q25).
- Section C comprises 6 Short Answer (SA) questions of 3 marks each (Q26 to Q31).
- Section D comprises 4 Long Answer (LA) questions of 5 marks each (Q32 to Q35).
- Section E comprises 3 Case-Based Integrated Units of Assessment of 4 marks each (Q36 to Q38) with sub-parts of values 1, 1 and 2 marks respectively. Internal choice is provided in the 2-mark question.
- Use of calculators is strictly prohibited. Use $\pi = 22/7$ wherever required unless stated otherwise.
SECTION A – MULTIPLE CHOICE QUESTIONS (1 Mark Each)
Q1.
If two positive integers $p$ and $q$ can be expressed as $p = ab^2$ and $q = a^3b$, where $a, b$ are prime numbers, then $\text{LCM}(p, q)$ is:
[1]
Q2.
If one of the zeroes of the quadratic polynomial $p(x) = (k – 1)x^2 + kx + 1$ is $-3$, then the value of $k$ is:
[1]
Q3.
The lines representing the linear equations $x – 2y = 0$ and $3x + 4y = 20$ intersect at the point:
[1]
Q4.
If $\frac{1}{2}$ is a root of the quadratic equation $x^2 + kx – \frac{5}{4} = 0$, then the value of $k$ is:
[1]
Q5.
Which term of the AP: $21, 18, 15, \dots$ is $-81$?
[1]
Q6.
If the distance between the points $A(x, 2)$ and $B(3, -6)$ is 10 units, then the values of $x$ are:
[1]
Q7.
If in $\triangle ABC$ and $\triangle DEF$, $\frac{AB}{DE} = \frac{BC}{FD}$, then the two triangles will be similar when:
[1]
Q8.
If $\cos \theta + \cos^2 \theta = 1$, then the value of $\sin^2 \theta + \sin^4 \theta$ is:
[1]
Q9.
If $x = 2\sin^2\theta$ and $y = 2\cos^2\theta + 1$, then the value of $x + y$ is:
[1]
Q10.
A kite is flying at a height of $60\text{ m}$ above the ground attached to a string inclined at $60^\circ$ to the horizontal. The length of the string is:
[1]
Q11.
A circle touches all four sides of a quadrilateral $ABCD$ whose sides are $AB = 6\text{ cm}$, $BC = 7\text{ cm}$, and $CD = 4\text{ cm}$. The length of $AD$ is:
[1]
Q12.
The area of the largest triangle that can be inscribed in a semicircle of radius $r$ is:
[1]
Q13.
Two cubes each of volume $64\text{ cm}^3$ are joined end to end. The total surface area of the resulting cuboid is:
[1]
Q14.
If the mode of a given data set is 45 and its mean is 27, then using the empirical relationship, its median is:
[1]
Q15.
A single unbiased standard die is thrown once. The probability of getting a prime number is:
[1]
Q16.
The class mark of the class interval $10 – 25$ is:
[1]
Q17.
A letter is chosen at random from the letters of the word ‘MATHEMATICS’. The probability that the chosen letter is a vowel is:
[1]
Q18.
The total number of factors of any prime number is:
[1]
Q19.
Assertion (A): The number $6^n$, where $n \in \mathbb{N}$, cannot end with the digit 0 for any natural number $n$.
Reason (R): The prime factorisation of $6^n$ is $(2 \times 3)^n$, which does not contain the prime factor 5. [1]
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Reason (R): The prime factorisation of $6^n$ is $(2 \times 3)^n$, which does not contain the prime factor 5. [1]
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Q20.
Assertion (A): The point $(-3, 0)$ lies on the negative direction of the x-axis.
Reason (R): The y-coordinate (ordinate) of any point lying on the x-axis is always zero. [1]
Reason (R): The y-coordinate (ordinate) of any point lying on the x-axis is always zero. [1]
SECTION B – VERY SHORT ANSWER QUESTIONS (2 Marks Each)
Q21.
Prove that $5 – \sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.
[2]
Q22.
In $\triangle ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $DE \parallel BC$. If $AD = 4\text{ cm}$, $DB = (x – 4)\text{ cm}$, $AE = 8\text{ cm}$, and $EC = (3x – 19)\text{ cm}$, find the value of $x$.
[2]
OR
Prove that the line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.
Q23.
Find the coordinates of a point $A$, where $AB$ is a diameter of a circle whose centre is $(2, -3)$ and $B$ is the point $(1, 4)$.
[2]
Q24.
Prove that:
$$(\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2 = 7 + \tan^2\theta + \cot^2\theta$$
[2]
OR
If $\sqrt{3}\sin\theta – \cos\theta = 0$ and $0^\circ < \theta < 90^\circ$, find the value of $\theta$ and hence evaluate $\sin^2\theta - \cos^2\theta$.
Q25.
Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.
[2]
SECTION C – SHORT ANSWER QUESTIONS (3 Marks Each)
Q26.
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = 2x^2 – 5x + 7$, find a quadratic polynomial whose zeroes are $2\alpha + 3\beta$ and $3\alpha + 2\beta$.
[3]
Q27.
Places $A$ and $B$ are $100\text{ km}$ apart on a highway. One car starts from $A$ and another from $B$ at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
[3]
OR
The cost of 5 pencils and 7 pens together is ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.
Q28.
Find the sum of all two-digit odd positive integers.
[3]
Q29.
Prove the trigonometric identity:
$$\frac{\tan\theta}{1 – \cot\theta} + \frac{\cot\theta}{1 – \tan\theta} = 1 + \sec\theta\csc\theta$$
[3]
Q30.
A triangle $ABC$ is drawn to circumscribe a circle of radius $4\text{ cm}$ such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8\text{ cm}$ and $6\text{ cm}$ respectively. Find the sides $AB$ and $AC$.
[3]
OR
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Q31.
One card is drawn from a well-shuffled deck of 52 playing cards. Find the probability of getting:
(a) A king of red colour
(b) A face card
(c) A spade or an ace [3]
(a) A king of red colour
(b) A face card
(c) A spade or an ace [3]
SECTION D – LONG ANSWER QUESTIONS (5 Marks Each)
Q32.
The sum of the areas of two squares is $468\text{ m}^2$. If the difference of their perimeters is $24\text{ m}$, find the sides of the two squares.
[5]
OR
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.
Q33.
State Basic Proportionality Theorem.
Using this theorem, prove that the diagonals of a trapezium divide each other proportionally. (i.e., In quadrilateral $ABCD$ with $AB \parallel DC$, diagonals $AC$ and $BD$ intersect at $O$, prove that $\frac{AO}{CO} = \frac{BO}{DO}$). [5]
Using this theorem, prove that the diagonals of a trapezium divide each other proportionally. (i.e., In quadrilateral $ABCD$ with $AB \parallel DC$, diagonals $AC$ and $BD$ intersect at $O$, prove that $\frac{AO}{CO} = \frac{BO}{DO}$). [5]
Q34.
A solid toy is in the form of a cylinder with hemispherical ends. The total length of the solid toy is $19\text{ cm}$ and the diameter of the cylinder is $7\text{ cm}$. Find the total volume and total surface area of the toy. (Take $\pi = 22/7$).
[5]
OR
A gulab jamun contains sugar syrup up to about $30\%$ of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with total length $5\text{ cm}$ and diameter $2.8\text{ cm}$. (Take $\pi = 22/7$).
Q35.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the Mode and Mean of this data:
$$\begin{array}{|c|c||c|c|} \hline \text{Number of students per teacher} & \text{Number of states/UTs} & \text{Number of students per teacher} & \text{Number of states/UTs} \\ \hline 15-20 & 3 & 35-40 & 3 \\ 20-25 & 8 & 40-45 & 0 \\ 25-30 & 9 & 45-50 & 0 \\ 30-35 & 10 & 50-55 & 2 \\ \hline \end{array}$$ [5]
$$\begin{array}{|c|c||c|c|} \hline \text{Number of students per teacher} & \text{Number of states/UTs} & \text{Number of students per teacher} & \text{Number of states/UTs} \\ \hline 15-20 & 3 & 35-40 & 3 \\ 20-25 & 8 & 40-45 & 0 \\ 25-30 & 9 & 45-50 & 0 \\ 30-35 & 10 & 50-55 & 2 \\ \hline \end{array}$$ [5]
SECTION E – CASE-BASED INTEGRATED UNITS (4 Marks Each)
Q36. Case Study 1 (Arithmetic Progression):
[4]
(a) Find the initial production in the 1st month ($a$). [1]
(b) Find the fixed monthly increment in production ($d$). [1]
(c) Find the total production of battery units in the first year (12 months). [2]
A manufacturing company produces electric scooter battery packs. Due to expanding market demand, the production increases uniformly by a fixed number of units every month. The company produced 1200 battery units in the 4th month and 1800 units in the 8th month.
Based on the above information, answer the following questions:
(a) Find the initial production in the 1st month ($a$). [1]
(b) Find the fixed monthly increment in production ($d$). [1]
(c) Find the total production of battery units in the first year (12 months). [2]
OR
In which month will the monthly production reach 3000 battery units? [2]
Q37. Case Study 2 (Coordinate Geometry):
[4]
(b) Find the coordinates of the midpoint of the line segment joining Station $A$ and Station $B$. [1]
(c) Calculate the perimeter of $\triangle ABC$ formed by the three stations. [2]
A city surveillance system maps three emergency response dispatch units on a Cartesian grid: Station $A(2, 3)$, Station $B(6, 7)$, and Station $C(8, 3)$, where 1 unit on the grid represents 1 kilometre.
(a) Find the direct straight-line distance between Station $A$ and Station $C$. [1]
(b) Find the coordinates of the midpoint of the line segment joining Station $A$ and Station $B$. [1]
(c) Calculate the perimeter of $\triangle ABC$ formed by the three stations. [2]
OR
If a central emergency command bunker $W$ is to be set up at the centroid of $\triangle ABC$, find the coordinates of bunker $W$. [2]
Q38. Case Study 3 (Some Applications of Trigonometry):
[4]
(b) Find the horizontal distance of the balloon from the surveyor at the first observation ($60^\circ$). [1]
(c) Calculate the distance travelled by the balloon during the 30-second interval, and find its horizontal speed in m/s. (Take $\sqrt{3} \approx 1.732$). [2]
A 1.2 m tall surveyor spots a weather observation balloon moving horizontally with the wind at a constant height above the ground. The angle of elevation of the balloon from the surveyor’s eye level at a given instant is $60^\circ$. After 30 seconds, the angle of elevation decreases to $30^\circ$. The balloon travels at a constant height of $88.2\text{ m}$ above the ground.
(a) Draw a neat mathematical labelled schematic diagram for this situation. [1]
(b) Find the horizontal distance of the balloon from the surveyor at the first observation ($60^\circ$). [1]
(c) Calculate the distance travelled by the balloon during the 30-second interval, and find its horizontal speed in m/s. (Take $\sqrt{3} \approx 1.732$). [2]
OR
Find the line-of-sight distance from the observer’s eye to the balloon at the second position ($30^\circ$). [2]
CBSE CLASS X MATHEMATICS (STANDARD) – SOLUTIONS & MARKING SCHEME (SET – 4)
SECTION A SOLUTIONS
Q1. (c) $a^3b^2$ [1 Mark]
Explanation: $\text{LCM}$ is the product of the highest power of each prime factor involved: $a^{\max(1,3)} \cdot b^{\max(2,1)} = a^3b^2$.
Explanation: $\text{LCM}$ is the product of the highest power of each prime factor involved: $a^{\max(1,3)} \cdot b^{\max(2,1)} = a^3b^2$.
Q2. (a) 4/3 [1 Mark]
Explanation: $p(-3) = (k – 1)(-3)^2 + k(-3) + 1 = 0 \implies 9k – 9 – 3k + 1 = 0 \implies 6k = 8 \implies k = \frac{4}{3}$.
Explanation: $p(-3) = (k – 1)(-3)^2 + k(-3) + 1 = 0 \implies 9k – 9 – 3k + 1 = 0 \implies 6k = 8 \implies k = \frac{4}{3}$.
Q3. (a) $(4, 2)$ [1 Mark]
Explanation: From $x – 2y = 0 \implies x = 2y$. Substituting in second: $3(2y) + 4y = 20 \implies 10y = 20 \implies y = 2, x = 4$.
Explanation: From $x – 2y = 0 \implies x = 2y$. Substituting in second: $3(2y) + 4y = 20 \implies 10y = 20 \implies y = 2, x = 4$.
Q4. (a) 2 [1 Mark]
Explanation: $\left(\frac{1}{2}\right)^2 + k\left(\frac{1}{2}\right) – \frac{5}{4} = 0 \implies \frac{1}{4} + \frac{k}{2} – \frac{5}{4} = 0 \implies \frac{k}{2} – 1 = 0 \implies k = 2$.
Explanation: $\left(\frac{1}{2}\right)^2 + k\left(\frac{1}{2}\right) – \frac{5}{4} = 0 \implies \frac{1}{4} + \frac{k}{2} – \frac{5}{4} = 0 \implies \frac{k}{2} – 1 = 0 \implies k = 2$.
Q5. (b) 35th [1 Mark]
Explanation: $a = 21, d = -3$. $a_n = -81 \implies 21 + (n – 1)(-3) = -81 \implies -3(n – 1) = -102 \implies n – 1 = 34 \implies n = 35$.
Explanation: $a = 21, d = -3$. $a_n = -81 \implies 21 + (n – 1)(-3) = -81 \implies -3(n – 1) = -102 \implies n – 1 = 34 \implies n = 35$.
Q6. (a) $9, -3$ [1 Mark]
Explanation: $\sqrt{(x – 3)^2 + (2 – (-6))^2} = 10 \implies (x – 3)^2 + 64 = 100 \implies (x – 3)^2 = 36 \implies x – 3 = \pm 6 \implies x = 9, -3$.
Explanation: $\sqrt{(x – 3)^2 + (2 – (-6))^2} = 10 \implies (x – 3)^2 + 64 = 100 \implies (x – 3)^2 = 36 \implies x – 3 = \pm 6 \implies x = 9, -3$.
Q7. (c) $\angle B = \angle D$ [1 Mark]
Explanation: In $\triangle ABC$ and $\triangle EFD$, the angle included between $AB$ and $BC$ is $\angle B$, and between $DE$ and $FD$ is $\angle D$. By SAS similarity, $\angle B = \angle D$.
Explanation: In $\triangle ABC$ and $\triangle EFD$, the angle included between $AB$ and $BC$ is $\angle B$, and between $DE$ and $FD$ is $\angle D$. By SAS similarity, $\angle B = \angle D$.
Q8. (c) 1 [1 Mark]
Explanation: $\cos\theta = 1 – \cos^2\theta = \sin^2\theta$. Then $\sin^2\theta + \sin^4\theta = \cos\theta + \cos^2\theta = 1$.
Explanation: $\cos\theta = 1 – \cos^2\theta = \sin^2\theta$. Then $\sin^2\theta + \sin^4\theta = \cos\theta + \cos^2\theta = 1$.
Q9. (b) 3 [1 Mark]
Explanation: $x + y = 2\sin^2\theta + 2\cos^2\theta + 1 = 2(\sin^2\theta + \cos^2\theta) + 1 = 2(1) + 1 = 3$.
Explanation: $x + y = 2\sin^2\theta + 2\cos^2\theta + 1 = 2(\sin^2\theta + \cos^2\theta) + 1 = 2(1) + 1 = 3$.
Q10. (a) $40\sqrt{3}\text{ m}$ [1 Mark]
Explanation: $\sin 60^\circ = \frac{\text{Height}}{\text{Length}} \implies \frac{\sqrt{3}}{2} = \frac{60}{L} \implies L = \frac{120}{\sqrt{3}} = 40\sqrt{3}\text{ m}$.
Explanation: $\sin 60^\circ = \frac{\text{Height}}{\text{Length}} \implies \frac{\sqrt{3}}{2} = \frac{60}{L} \implies L = \frac{120}{\sqrt{3}} = 40\sqrt{3}\text{ m}$.
Q11. (b) 3 cm [1 Mark]
Explanation: $AB + CD = AD + BC \implies 6 + 4 = AD + 7 \implies AD = 10 – 7 = 3\text{ cm}$.
Explanation: $AB + CD = AD + BC \implies 6 + 4 = AD + 7 \implies AD = 10 – 7 = 3\text{ cm}$.
Q12. (a) $r^2$ [1 Mark]
Explanation: The base of the largest triangle is the diameter $2r$ and its height is the radius $r$. $\text{Area} = \frac{1}{2} \times 2r \times r = r^2$.
Explanation: The base of the largest triangle is the diameter $2r$ and its height is the radius $r$. $\text{Area} = \frac{1}{2} \times 2r \times r = r^2$.
Q13. (b) $160\text{ cm}^2$ [1 Mark]
Explanation: Side of each cube $a = \sqrt[3]{64} = 4\text{ cm}$. Resulting cuboid: $l = 8\text{ cm}, b = 4\text{ cm}, h = 4\text{ cm}$.
$\text{TSA} = 2(lb + bh + hl) = 2(32 + 16 + 32) = 2(80) = 160\text{ cm}^2$.
Explanation: Side of each cube $a = \sqrt[3]{64} = 4\text{ cm}$. Resulting cuboid: $l = 8\text{ cm}, b = 4\text{ cm}, h = 4\text{ cm}$.
$\text{TSA} = 2(lb + bh + hl) = 2(32 + 16 + 32) = 2(80) = 160\text{ cm}^2$.
Q14. (a) 33 [1 Mark]
Explanation: $\text{Mode} = 3\text{Median} – 2\text{Mean} \implies 45 = 3\text{Median} – 2(27) \implies 3\text{Median} = 99 \implies \text{Median} = 33$.
Explanation: $\text{Mode} = 3\text{Median} – 2\text{Mean} \implies 45 = 3\text{Median} – 2(27) \implies 3\text{Median} = 99 \implies \text{Median} = 33$.
Q15. (b) 1/2 [1 Mark]
Explanation: Prime numbers on a die are $\{2, 3, 5\}$ (3 outcomes). $P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}$.
Explanation: Prime numbers on a die are $\{2, 3, 5\}$ (3 outcomes). $P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}$.
Q16. (b) 17.5 [1 Mark]
Explanation: $\text{Class mark} = \frac{10 + 25}{2} = \frac{35}{2} = 17.5$.
Explanation: $\text{Class mark} = \frac{10 + 25}{2} = \frac{35}{2} = 17.5$.
Q17. (c) 4/11 [1 Mark]
Explanation: Total letters in ‘MATHEMATICS’ $= 11$. Vowels are $\{A, E, A, I\}$ (4 vowels). $P = \frac{4}{11}$.
Explanation: Total letters in ‘MATHEMATICS’ $= 11$. Vowels are $\{A, E, A, I\}$ (4 vowels). $P = \frac{4}{11}$.
Q18. (b) 2 [1 Mark]
Explanation: A prime number has exactly two factors: 1 and the number itself.
Explanation: A prime number has exactly two factors: 1 and the number itself.
Q19. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A). [1 Mark]
Explanation: $6^n = (2 \times 3)^n$. By Fundamental Theorem of Arithmetic, the uniqueness of prime factorisation ensures there is no factor 5, so it cannot end in 0.
Explanation: $6^n = (2 \times 3)^n$. By Fundamental Theorem of Arithmetic, the uniqueness of prime factorisation ensures there is no factor 5, so it cannot end in 0.
Q20. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A). [1 Mark]
Explanation: Points with $y = 0$ and $x < 0$ lie on the negative x-axis.
Explanation: Points with $y = 0$ and $x < 0$ lie on the negative x-axis.
SECTION B SOLUTIONS
Q21. Proof of Irrationality: [2 Marks]
- Let $5 – \sqrt{3}$ be a rational number equal to $\frac{a}{b}$, where $a, b$ are coprime integers and $b \neq 0$. [0.5 Mark]
- $\sqrt{3} = 5 – \frac{a}{b} = \frac{5b – a}{b}$. [0.5 Mark]
- Since $a$ and $b$ are integers, $\frac{5b – a}{b}$ is rational, implying $\sqrt{3}$ is rational. [0.5 Mark]
- This contradicts the given fact that $\sqrt{3}$ is irrational. Hence, $5 – \sqrt{3}$ is irrational. [0.5 Mark]
Q22. Finding $x$: [2 Marks]
By Basic Proportionality Theorem: $\frac{AD}{DB} = \frac{AE}{EC}$ [0.5 Mark]
$\frac{4}{x – 4} = \frac{8}{3x – 19} \implies \frac{1}{x – 4} = \frac{2}{3x – 19}$ [0.5 Mark]
$3x – 19 = 2(x – 4) \implies 3x – 19 = 2x – 8 \implies \mathbf{x = 11\text{ cm}}$. [1 Mark]
OR
Let $D$ be midpoint of $AB$ ($AD = DB$). Line through $D$ parallel to $BC$ meets $AC$ at $E$.
By BPT: $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{1}{1} = \frac{AE}{EC} \implies AE = EC$. Hence $E$ bisects $AC$. [2 Marks]
By Basic Proportionality Theorem: $\frac{AD}{DB} = \frac{AE}{EC}$ [0.5 Mark]
$\frac{4}{x – 4} = \frac{8}{3x – 19} \implies \frac{1}{x – 4} = \frac{2}{3x – 19}$ [0.5 Mark]
$3x – 19 = 2(x – 4) \implies 3x – 19 = 2x – 8 \implies \mathbf{x = 11\text{ cm}}$. [1 Mark]
OR
Let $D$ be midpoint of $AB$ ($AD = DB$). Line through $D$ parallel to $BC$ meets $AC$ at $E$.
By BPT: $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{1}{1} = \frac{AE}{EC} \implies AE = EC$. Hence $E$ bisects $AC$. [2 Marks]
Q23. Coordinates of Point A: [2 Marks]
Let $A(x, y)$. Centre $C(2, -3)$ is the midpoint of diameter $AB$ with $B(1, 4)$.
$\frac{x + 1}{2} = 2 \implies x + 1 = 4 \implies \mathbf{x = 3}$ [1 Mark]
$\frac{y + 4}{2} = -3 \implies y + 4 = -6 \implies \mathbf{y = -10}$ [0.5 Mark]
Coordinates of point $A$ are $\mathbf{(3, -10)}$. [0.5 Mark]
Let $A(x, y)$. Centre $C(2, -3)$ is the midpoint of diameter $AB$ with $B(1, 4)$.
$\frac{x + 1}{2} = 2 \implies x + 1 = 4 \implies \mathbf{x = 3}$ [1 Mark]
$\frac{y + 4}{2} = -3 \implies y + 4 = -6 \implies \mathbf{y = -10}$ [0.5 Mark]
Coordinates of point $A$ are $\mathbf{(3, -10)}$. [0.5 Mark]
Q24. Trigonometric Identity Proof: [2 Marks]
$\text{LHS} = (\sin^2\theta + \csc^2\theta + 2\sin\theta\csc\theta) + (\cos^2\theta + \sec^2\theta + 2\cos\theta\sec\theta)$
$= (\sin^2\theta + \cos^2\theta) + (1 + \cot^2\theta) + 2(1) + (1 + \tan^2\theta) + 2(1)$ [1 Mark]
$= 1 + 1 + \cot^2\theta + 2 + 1 + \tan^2\theta + 2 = \mathbf{7 + \tan^2\theta + \cot^2\theta} = \text{RHS}$. [1 Mark]
OR
$\sqrt{3}\sin\theta = \cos\theta \implies \tan\theta = \frac{1}{\sqrt{3}} \implies \mathbf{\theta = 30^\circ}$. [1 Mark]
$\sin^2 30^\circ – \cos^2 30^\circ = \left(\frac{1}{2}\right)^2 – \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} – \frac{3}{4} = \mathbf{-\frac{1}{2}}$. [1 Mark]
$\text{LHS} = (\sin^2\theta + \csc^2\theta + 2\sin\theta\csc\theta) + (\cos^2\theta + \sec^2\theta + 2\cos\theta\sec\theta)$
$= (\sin^2\theta + \cos^2\theta) + (1 + \cot^2\theta) + 2(1) + (1 + \tan^2\theta) + 2(1)$ [1 Mark]
$= 1 + 1 + \cot^2\theta + 2 + 1 + \tan^2\theta + 2 = \mathbf{7 + \tan^2\theta + \cot^2\theta} = \text{RHS}$. [1 Mark]
OR
$\sqrt{3}\sin\theta = \cos\theta \implies \tan\theta = \frac{1}{\sqrt{3}} \implies \mathbf{\theta = 30^\circ}$. [1 Mark]
$\sin^2 30^\circ – \cos^2 30^\circ = \left(\frac{1}{2}\right)^2 – \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} – \frac{3}{4} = \mathbf{-\frac{1}{2}}$. [1 Mark]
Q25. Concentric Circles Chord Bisected: [2 Marks]
Let chord $AB$ of larger circle touch smaller circle at $P$. Centre is $O$.
$AB$ is tangent to smaller circle at $P \implies OP \perp AB$. [1 Mark]
$AB$ is a chord of the larger circle and $OP \perp AB$. Since the perpendicular from the centre to a chord bisects the chord, $\mathbf{AP = PB}$. Hence proved. [1 Mark]
Let chord $AB$ of larger circle touch smaller circle at $P$. Centre is $O$.
$AB$ is tangent to smaller circle at $P \implies OP \perp AB$. [1 Mark]
$AB$ is a chord of the larger circle and $OP \perp AB$. Since the perpendicular from the centre to a chord bisects the chord, $\mathbf{AP = PB}$. Hence proved. [1 Mark]
SECTION C SOLUTIONS
Q26. Quadratic Polynomial Formation: [3 Marks]
$p(x) = 2x^2 – 5x + 7 \implies \alpha + \beta = \frac{5}{2}$, $\alpha\beta = \frac{7}{2}$. [0.5 Mark]
Let new zeroes be $S_1 = 2\alpha + 3\beta$ and $S_2 = 3\alpha + 2\beta$.
$\text{Sum of new zeroes} = 5(\alpha + \beta) = 5\left(\frac{5}{2}\right) = \frac{25}{2}$. [1 Mark]
$\text{Product} = (2\alpha + 3\beta)(3\alpha + 2\beta) = 6\alpha^2 + 13\alpha\beta + 6\beta^2 = 6(\alpha + \beta)^2 + \alpha\beta = 6\left(\frac{25}{4}\right) + \frac{7}{2} = \frac{75}{2} + \frac{7}{2} = 41$. [1 Mark]
Polynomial is $k\left(x^2 – \frac{25}{2}x + 41\right) \implies \mathbf{2x^2 – 25x + 82}$. [0.5 Mark]
$p(x) = 2x^2 – 5x + 7 \implies \alpha + \beta = \frac{5}{2}$, $\alpha\beta = \frac{7}{2}$. [0.5 Mark]
Let new zeroes be $S_1 = 2\alpha + 3\beta$ and $S_2 = 3\alpha + 2\beta$.
$\text{Sum of new zeroes} = 5(\alpha + \beta) = 5\left(\frac{5}{2}\right) = \frac{25}{2}$. [1 Mark]
$\text{Product} = (2\alpha + 3\beta)(3\alpha + 2\beta) = 6\alpha^2 + 13\alpha\beta + 6\beta^2 = 6(\alpha + \beta)^2 + \alpha\beta = 6\left(\frac{25}{4}\right) + \frac{7}{2} = \frac{75}{2} + \frac{7}{2} = 41$. [1 Mark]
Polynomial is $k\left(x^2 – \frac{25}{2}x + 41\right) \implies \mathbf{2x^2 – 25x + 82}$. [0.5 Mark]
Q27. Speed of Two Cars: [3 Marks]
Let speed of car from $A$ be $u\text{ km/h}$ and from $B$ be $v\text{ km/h}$ ($u > v$).
Same direction: $5(u – v) = 100 \implies u – v = 20$ …(1) [1 Mark]
Opposite directions: $1(u + v) = 100 \implies u + v = 100$ …(2) [1 Mark]
Adding (1) and (2): $2u = 120 \implies \mathbf{u = 60\text{ km/h}}$; from (2): $\mathbf{v = 40\text{ km/h}}$. [1 Mark]
OR
$5x + 7y = 50$ …(1); $7x + 5y = 46$ …(2)
Adding: $12x + 12y = 96 \implies x + y = 8$ …(3) [1 Mark]
Subtracting: $-2x + 2y = 4 \implies -x + y = 2$ …(4) [1 Mark]
Solving (3) and (4): $\mathbf{y = 5\text{ (Pen = ₹5)}}, \mathbf{x = 3\text{ (Pencil = ₹3)}}$. [1 Mark]
Let speed of car from $A$ be $u\text{ km/h}$ and from $B$ be $v\text{ km/h}$ ($u > v$).
Same direction: $5(u – v) = 100 \implies u – v = 20$ …(1) [1 Mark]
Opposite directions: $1(u + v) = 100 \implies u + v = 100$ …(2) [1 Mark]
Adding (1) and (2): $2u = 120 \implies \mathbf{u = 60\text{ km/h}}$; from (2): $\mathbf{v = 40\text{ km/h}}$. [1 Mark]
OR
$5x + 7y = 50$ …(1); $7x + 5y = 46$ …(2)
Adding: $12x + 12y = 96 \implies x + y = 8$ …(3) [1 Mark]
Subtracting: $-2x + 2y = 4 \implies -x + y = 2$ …(4) [1 Mark]
Solving (3) and (4): $\mathbf{y = 5\text{ (Pen = ₹5)}}, \mathbf{x = 3\text{ (Pencil = ₹3)}}$. [1 Mark]
Q28. Sum of Two-Digit Odd Numbers: [3 Marks]
Sequence: $11, 13, 15, \dots, 99$. Here $a = 11, d = 2, l = 99$. [1 Mark]
$99 = 11 + (n – 1)2 \implies 88 = 2(n – 1) \implies n – 1 = 44 \implies n = 45$. [1 Mark]
$S_{45} = \frac{45}{2}(a + l) = \frac{45}{2}(11 + 99) = \frac{45}{2}(110) = 45 \times 55 = \mathbf{2475}$. [1 Mark]
Sequence: $11, 13, 15, \dots, 99$. Here $a = 11, d = 2, l = 99$. [1 Mark]
$99 = 11 + (n – 1)2 \implies 88 = 2(n – 1) \implies n – 1 = 44 \implies n = 45$. [1 Mark]
$S_{45} = \frac{45}{2}(a + l) = \frac{45}{2}(11 + 99) = \frac{45}{2}(110) = 45 \times 55 = \mathbf{2475}$. [1 Mark]
Q29. Trigonometric Identity: [3 Marks]
$\text{LHS} = \frac{\frac{\sin\theta}{\cos\theta}}{1 – \frac{\cos\theta}{\sin\theta}} + \frac{\frac{\cos\theta}{\sin\theta}}{1 – \frac{\sin\theta}{\cos\theta}} = \frac{\sin^2\theta}{\cos\theta(\sin\theta – \cos\theta)} – \frac{\cos^2\theta}{\sin\theta(\sin\theta – \cos\theta)}$ [1 Mark]
$= \frac{\sin^3\theta – \cos^3\theta}{\sin\theta\cos\theta(\sin\theta – \cos\theta)} = \frac{(\sin\theta – \cos\theta)(\sin^2\theta + \cos^2\theta + \sin\theta\cos\theta)}{\sin\theta\cos\theta(\sin\theta – \cos\theta)}$ [1 Mark]
$= \frac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} + 1 = \mathbf{1 + \sec\theta\csc\theta} = \text{RHS}$. Hence Proved. [1 Mark]
$\text{LHS} = \frac{\frac{\sin\theta}{\cos\theta}}{1 – \frac{\cos\theta}{\sin\theta}} + \frac{\frac{\cos\theta}{\sin\theta}}{1 – \frac{\sin\theta}{\cos\theta}} = \frac{\sin^2\theta}{\cos\theta(\sin\theta – \cos\theta)} – \frac{\cos^2\theta}{\sin\theta(\sin\theta – \cos\theta)}$ [1 Mark]
$= \frac{\sin^3\theta – \cos^3\theta}{\sin\theta\cos\theta(\sin\theta – \cos\theta)} = \frac{(\sin\theta – \cos\theta)(\sin^2\theta + \cos^2\theta + \sin\theta\cos\theta)}{\sin\theta\cos\theta(\sin\theta – \cos\theta)}$ [1 Mark]
$= \frac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} + 1 = \mathbf{1 + \sec\theta\csc\theta} = \text{RHS}$. Hence Proved. [1 Mark]
Q30. Circumscribed Triangle Sides: [3 Marks]
Let $AF = AE = x$. Sides are $a = 14, b = x + 6, c = x + 8$.
Semi-perimeter $s = \frac{14 + x + 6 + x + 8}{2} = x + 14$. [0.5 Mark]
$\text{Area}(\triangle ABC) = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{(x + 14)(x)(8)(6)} = \sqrt{48x(x + 14)}$. [1 Mark]
Also $\text{Area} = r \times s = 4(x + 14)$.
Equating: $48x(x + 14) = 16(x + 14)^2 \implies 3x = x + 14 \implies 2x = 14 \implies x = 7\text{ cm}$. [1 Mark]
$\mathbf{AB = 7 + 8 = 15\text{ cm}}$ and $\mathbf{AC = 7 + 6 = 13\text{ cm}}$. [0.5 Mark]
OR
Diagram and construction joining vertices and contact points to centre $O$. [1 Mark]
Proving congruent pairs of triangles at centre subtend equal angles. [1 Mark]
Sum of all angles $= 360^\circ \implies 2(\angle AOB + \angle COD) = 360^\circ \implies \mathbf{\angle AOB + \angle COD = 180^\circ}$. [1 Mark]
Let $AF = AE = x$. Sides are $a = 14, b = x + 6, c = x + 8$.
Semi-perimeter $s = \frac{14 + x + 6 + x + 8}{2} = x + 14$. [0.5 Mark]
$\text{Area}(\triangle ABC) = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{(x + 14)(x)(8)(6)} = \sqrt{48x(x + 14)}$. [1 Mark]
Also $\text{Area} = r \times s = 4(x + 14)$.
Equating: $48x(x + 14) = 16(x + 14)^2 \implies 3x = x + 14 \implies 2x = 14 \implies x = 7\text{ cm}$. [1 Mark]
$\mathbf{AB = 7 + 8 = 15\text{ cm}}$ and $\mathbf{AC = 7 + 6 = 13\text{ cm}}$. [0.5 Mark]
OR
Diagram and construction joining vertices and contact points to centre $O$. [1 Mark]
Proving congruent pairs of triangles at centre subtend equal angles. [1 Mark]
Sum of all angles $= 360^\circ \implies 2(\angle AOB + \angle COD) = 360^\circ \implies \mathbf{\angle AOB + \angle COD = 180^\circ}$. [1 Mark]
Q31. Card Probability: [3 Marks]
Total cards $= 52$.
Total cards $= 52$.
- (a) Red kings (Heart and Diamond) $= 2 \implies P = \frac{2}{52} = \mathbf{\frac{1}{26}}$. [1 Mark]
- (b) Face cards (4 Jacks, 4 Queens, 4 Kings) $= 12 \implies P = \frac{12}{52} = \mathbf{\frac{3}{13}}$. [1 Mark]
- (c) Spades (13) + non-spade Aces (3) $= 16 \implies P = \frac{16}{52} = \mathbf{\frac{4}{13}}$. [1 Mark]
SECTION D SOLUTIONS
Q32. Two Squares Problem: [5 Marks]
Let sides of squares be $x\text{ m}$ and $y\text{ m}$ ($x > y$).
$x^2 + y^2 = 468$ …(1)
$4x – 4y = 24 \implies x – y = 6 \implies x = y + 6$ …(2) [1.5 Marks]
Substituting (2) into (1): $(y + 6)^2 + y^2 = 468 \implies 2y^2 + 12y + 36 = 468$
$2y^2 + 12y – 432 = 0 \implies y^2 + 6y – 216 = 0$ [1.5 Marks]
$(y + 18)(y – 12) = 0 \implies y = 12\text{ m}$ (side cannot be negative).
$x = 12 + 6 = 18\text{ m}$. Sides are $\mathbf{18\text{ m}}$ and $\mathbf{12\text{ m}}$. [2 Marks]
OR
Let number of articles be $x$. Cost of each article $= 2x + 3$.
$x(2x + 3) = 90 \implies 2x^2 + 3x – 90 = 0$ [2 Marks]
$2x^2 + 15x – 12x – 90 = 0 \implies x(2x + 15) – 6(2x + 15) = 0 \implies x = 6$. [1.5 Marks]
Number of articles $= \mathbf{6}$, Cost of each article $= 2(6) + 3 = \mathbf{₹15}$. [1.5 Marks]
Let sides of squares be $x\text{ m}$ and $y\text{ m}$ ($x > y$).
$x^2 + y^2 = 468$ …(1)
$4x – 4y = 24 \implies x – y = 6 \implies x = y + 6$ …(2) [1.5 Marks]
Substituting (2) into (1): $(y + 6)^2 + y^2 = 468 \implies 2y^2 + 12y + 36 = 468$
$2y^2 + 12y – 432 = 0 \implies y^2 + 6y – 216 = 0$ [1.5 Marks]
$(y + 18)(y – 12) = 0 \implies y = 12\text{ m}$ (side cannot be negative).
$x = 12 + 6 = 18\text{ m}$. Sides are $\mathbf{18\text{ m}}$ and $\mathbf{12\text{ m}}$. [2 Marks]
OR
Let number of articles be $x$. Cost of each article $= 2x + 3$.
$x(2x + 3) = 90 \implies 2x^2 + 3x – 90 = 0$ [2 Marks]
$2x^2 + 15x – 12x – 90 = 0 \implies x(2x + 15) – 6(2x + 15) = 0 \implies x = 6$. [1.5 Marks]
Number of articles $= \mathbf{6}$, Cost of each article $= 2(6) + 3 = \mathbf{₹15}$. [1.5 Marks]
Q33. BPT and Trapezium Application: [5 Marks]
- BPT Statement & Proof: [2.5 Marks]
- Trapezium Application: In trapezium $ABCD$ with $AB \parallel DC$, draw line $OE \parallel AB$ meeting $AD$ at $E$.
In $\triangle ABD$: $\frac{AE}{ED} = \frac{BO}{OD}$ …(1) (by BPT) [1 Mark]
In $\triangle ADC$: $\frac{AE}{ED} = \frac{AO}{OC}$ …(2) (by BPT) [1 Mark]
From (1) and (2): $\mathbf{\frac{AO}{OC} = \frac{BO}{OD} \implies \frac{AO}{BO} = \frac{CO}{DO}}$. Hence Proved. [0.5 Mark]
Q34. Toy Volume and Surface Area: [5 Marks]
Radius $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$. Height of cylinder $h = 19 – 2(3.5) = 12\text{ cm}$.
$\text{Volume} = \pi r^2 h + 2\left(\frac{2}{3}\pi r^3\right) = \pi r^2\left(h + \frac{4}{3}r\right) = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times \left[12 + \frac{4}{3}\left(\frac{7}{2}\right)\right]$
$= \frac{77}{2} \times \frac{50}{3} = \frac{1925}{3} \approx \mathbf{641.67\text{ cm}^3}$. [2.5 Marks]
$\text{Total Surface Area} = 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r) = 2 \times \frac{22}{7} \times \frac{7}{2} \times (12 + 7) = 22 \times 19 = \mathbf{418\text{ cm}^2}$. [2.5 Marks]
OR
$r = 1.4\text{ cm}$, Cylinder height $h = 5 – 2(1.4) = 2.2\text{ cm}$.
Volume of 1 gulab jamun $= \pi r^2 h + \frac{4}{3}\pi r^3 = \frac{22}{7} \times (1.4)^2 \times \left[2.2 + \frac{4}{3}(1.4)\right] = 6.16 \times 4.067 \approx 25.05\text{ cm}^3$. [2.5 Marks]
Volume of 45 gulab jamuns $= 45 \times 25.05 = 1127.25\text{ cm}^3$. [1 Mark]
$\text{Quantity of syrup} = 30\% \times 1127.25 \approx \mathbf{338\text{ cm}^3}$. [1.5 Marks]
Radius $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$. Height of cylinder $h = 19 – 2(3.5) = 12\text{ cm}$.
$\text{Volume} = \pi r^2 h + 2\left(\frac{2}{3}\pi r^3\right) = \pi r^2\left(h + \frac{4}{3}r\right) = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times \left[12 + \frac{4}{3}\left(\frac{7}{2}\right)\right]$
$= \frac{77}{2} \times \frac{50}{3} = \frac{1925}{3} \approx \mathbf{641.67\text{ cm}^3}$. [2.5 Marks]
$\text{Total Surface Area} = 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r) = 2 \times \frac{22}{7} \times \frac{7}{2} \times (12 + 7) = 22 \times 19 = \mathbf{418\text{ cm}^2}$. [2.5 Marks]
OR
$r = 1.4\text{ cm}$, Cylinder height $h = 5 – 2(1.4) = 2.2\text{ cm}$.
Volume of 1 gulab jamun $= \pi r^2 h + \frac{4}{3}\pi r^3 = \frac{22}{7} \times (1.4)^2 \times \left[2.2 + \frac{4}{3}(1.4)\right] = 6.16 \times 4.067 \approx 25.05\text{ cm}^3$. [2.5 Marks]
Volume of 45 gulab jamuns $= 45 \times 25.05 = 1127.25\text{ cm}^3$. [1 Mark]
$\text{Quantity of syrup} = 30\% \times 1127.25 \approx \mathbf{338\text{ cm}^3}$. [1.5 Marks]
Q35. Mode and Mean: [5 Marks]
Modal class is $30 – 35$ ($l = 30, f_1 = 10, f_0 = 9, f_2 = 3, h = 5$).
$\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h = 30 + \left(\frac{10 – 9}{20 – 9 – 3}\right) \times 5 = 30 + \frac{5}{8} = \mathbf{30.63}$. [2.5 Marks]
$\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3(17.5) + 8(22.5) + 9(27.5) + 10(32.5) + 3(37.5) + 2(52.5)}{35} = \frac{1022.5}{35} \approx \mathbf{29.21}$. [2.5 Marks]
Modal class is $30 – 35$ ($l = 30, f_1 = 10, f_0 = 9, f_2 = 3, h = 5$).
$\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h = 30 + \left(\frac{10 – 9}{20 – 9 – 3}\right) \times 5 = 30 + \frac{5}{8} = \mathbf{30.63}$. [2.5 Marks]
$\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3(17.5) + 8(22.5) + 9(27.5) + 10(32.5) + 3(37.5) + 2(52.5)}{35} = \frac{1022.5}{35} \approx \mathbf{29.21}$. [2.5 Marks]
SECTION E SOLUTIONS
Q36. Case Study 1: [4 Marks]
- (a) $a + 3d = 1200, a + 7d = 1800 \implies 4d = 600 \implies d = 150$.
$a = 1200 – 3(150) = \mathbf{750\text{ units}}$. [1 Mark] - (b) Fixed monthly increase $d = \mathbf{150\text{ units}}$. [1 Mark]
- (c) $S_{12} = \frac{12}{2}[2(750) + 11(150)] = 6[1500 + 1650] = 6[3150] = \mathbf{18,900\text{ units}}$. [2 Marks]
Q37. Case Study 2: [4 Marks]
- (a) $AC = \sqrt{(8 – 2)^2 + (3 – 3)^2} = \sqrt{6^2 + 0} = \mathbf{6\text{ km}}$. [1 Mark]
- (b) $\text{Midpoint of } AB = \left(\frac{2 + 6}{2}, \frac{3 + 7}{2}\right) = \mathbf{(4, 5)}$. [1 Mark]
- (c) $AB = \sqrt{(6-2)^2 + (7-3)^2} = \sqrt{16+16} = 4\sqrt{2}\text{ km}$; $BC = \sqrt{(8-6)^2 + (3-7)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}\text{ km}$.
$\text{Perimeter} = 6 + 4\sqrt{2} + 2\sqrt{5} \approx 6 + 5.66 + 4.47 = \mathbf{16.13\text{ km}}$. [2 Marks]
Q38. Case Study 3: [4 Marks]
- (a) Labelled diagram showing eye-level height $h = 88.2 – 1.2 = 87\text{ m}$, with angles $60^\circ$ and $30^\circ$. [1 Mark]
- (b) In $\triangle$ with $60^\circ$: $\tan 60^\circ = \frac{87}{x} \implies \sqrt{3} = \frac{87}{x} \implies x = \frac{87}{\sqrt{3}} = \mathbf{29\sqrt{3}\text{ m}} \approx 50.23\text{ m}$. [1 Mark]
- (c) In $\triangle$ with $30^\circ$: $\tan 30^\circ = \frac{87}{y} \implies \frac{1}{\sqrt{3}} = \frac{87}{y} \implies y = 87\sqrt{3}\text{ m}$.
$\text{Distance} = y – x = 58\sqrt{3}\text{ m} = 58(1.732) \approx \mathbf{100.46\text{ m}}$.
$\text{Speed} = \frac{58\sqrt{3}}{30}\text{ m/s} = \mathbf{3.35\text{ m/s}}$ (or $12.06\text{ km/h}$). [2 Marks]